I am confused by eq. 1 and 2. You converted (1) from cartesian to "wonky" cartesian. You have the equation of the normal line as y=x, which is correct, but that second equation seems to come out of nowhere.

What does the original function actually represent? If you graph the curve, and y=x (the normal line) the answers become more clear.

I am confused by eq. 1 and 2. You converted (1) from cartesian to "wonky" cartesian. You have the equation of the normal line as y=x, which is correct, but that second equation seems to come out of nowhere.

What does the original function actually represent? If you graph the curve, and y=x (the normal line) the answers become more clear.

As a side note, limiting the domain might be a bad idea ;-)

ok , so how do i get the equation of the curve given its parametric equation ? I tried to eliminate the parameters and got that second equation .

Well, the curve given is a circle with radius 1.
This can be verified: The equation of a circle is
x2+y2=r2
since you were given:
x=cos(b)
y=sin(b)
x2+y2=r2 goes to:
cos2(b)+sin2(b)=r2
1=r2
1=r

If you draw the circle and the line y=x, which according to the problem is:
sin(b)=cos(b)
sin(b)/cos(b)=1
tan(b)=1

As for the intersections:
tan(b)=1=x2+y2 I think you had zero here by mistake.
If you check your unit circle, this occurs at Pi/4 +kPi, where k is an integer. (This works for all real domains, not just zero to 2Pi)

So you were on the right path from the get go, but the extra transform threw things off.

Well, the curve given is a circle with radius 1.
This can be verified: The equation of a circle is
x2+y2=r2
since you were given:
x=cos(b)
y=sin(b)
x2+y2=r2 goes to:
cos2(b)+sin2(b)=r2
1=r2
1=r

If you draw the circle and the line y=x, which according to the problem is:
sin(b)=cos(b)
sin(b)/cos(b)=1
tan(b)=1

As for the intersections:
tan(b)=1=x2+y2 I think you had zero here by mistake.
If you check your unit circle, this occurs at Pi/4 +kPi, where k is an integer. (This works for all real domains, not just zero to 2Pi)

So you were on the right path from the get go, but the extra transform threw things off.

Staff: Mentor

Doesn't matter if it's high school or college or whatever. Calculus questions should go to the Calculus & Beyond section.

The line y = x comes about because the problem asks for the normal on the parametric curve at b = pi/4. This point corresponds to the point (sqrt(2)/2, sqrt(2)/2) on the circle. Because the curve is a circle, the normal to the curve intersects the circle at a point directly across the circle, at (-sqrt(2)/2, -sqrt(2)/2). The x and y values in these points are the coordinates.